Question 336365
{{{2x^2-2x-1=0}}} Start with the given equation.



Notice that the quadratic {{{2x^2-2x-1}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=2}}}, {{{B=-2}}}, and {{{C=-1}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-2) +- sqrt( (-2)^2-4(2)(-1) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=-2}}}, and {{{C=-1}}}



{{{x = (2 +- sqrt( (-2)^2-4(2)(-1) ))/(2(2))}}} Negate {{{-2}}} to get {{{2}}}. 



{{{x = (2 +- sqrt( 4-4(2)(-1) ))/(2(2))}}} Square {{{-2}}} to get {{{4}}}. 



{{{x = (2 +- sqrt( 4--8 ))/(2(2))}}} Multiply {{{4(2)(-1)}}} to get {{{-8}}}



{{{x = (2 +- sqrt( 4+8 ))/(2(2))}}} Rewrite {{{sqrt(4--8)}}} as {{{sqrt(4+8)}}}



{{{x = (2 +- sqrt( 12 ))/(2(2))}}} Add {{{4}}} to {{{8}}} to get {{{12}}}



{{{x = (2 +- sqrt( 12 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (2 +- 2*sqrt(3))/(4)}}} Simplify the square root  (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)  



{{{x = (2+2*sqrt(3))/(4)}}} or {{{x = (2-2*sqrt(3))/(4)}}} Break up the expression.



{{{x = (1+sqrt(3))/(2)}}} or {{{x = (1-sqrt(3))/(2)}}} Reduce.  



So the solutions are {{{x = (1+sqrt(3))/(2)}}} or {{{x = (1-sqrt(3))/(2)}}} 



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Jim